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The Speed of Sound (Trick Question?)

  1. Jan 20, 2005 #1
    I have a seemingly easy question I think.......but wanted to run it by this board before I submitted my answer.

    Homework Question
    The sound produced by the loudspeaker in the (attached) drawing has a frequency of 10700 Hz and arrives at the microphone via two different paths. The sound travels through the left tube LXM, which has a fixed length. Simultaneously, the sound travels through the right tube LYM, the length of which can be changed by moving the sliding section. At M, the sound waves coming from the two paths interfere. As the length of the path LYM is changed, the sound loudness detected by the microphone changes. When the sliding section is pulled out by 0.0290 m, the loudness changes from a maximum to a minimum. Find the speed at which sound travels through the gas in the tube.


    Is this a trick question? Isn't the speed of sound always 343 m/s?

    Thank you in advance!
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2005 #2

    dextercioby

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    No,the speed of sound varies in large,very large limits.From 6Km/s in graptite to very small speeds in rarefied gases...

    Daniel.
     
  4. Jan 20, 2005 #3

    dextercioby

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    HINT:The sound waves are stationary...What is the condition relating consecutive minima and maxima for a stationary wave...??

    Daniel.

    P.S.It involves the wavelength...
     
  5. Jan 20, 2005 #4
    Hmmmmmm,

    The condition relating consecutive minima and maxima for a stationary wave would be pressure right? The speed of a wave is dependent upon the medium in which it exists and in this case, the medium is simply air, so standard air pressure I would imagine would factor into the equation. Am I on the right track? I also know that {Velocity = (frequency)(wavelength)}. The longer the wavelength the faster the speed? I am given the frequency but I am one variable too short to be able to use tha above equation.

    This seems like such an easy question but is stumping me big time!
     
  6. Jan 20, 2005 #5

    dextercioby

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    You know too much... :tongue2: to get this problem solved...You didn't answer my question,though...

    I'll let u think about it.I'll give another
    HINT:Think about the waves on a string...How do you relate consecutive minima & maxima to the wavelength??

    Daniel.
     
  7. Jan 21, 2005 #6
    Hi Daniel,

    Thank you again for your help!

    To maybe answer your question this time........

    The distance between consecutive maxima or consecutive minima is the wavelength of the sound. The wavelength is generated by the pulse, a compression that propagates away from the source of the sound.

    I can calculate the wavelength if I know the frequency (given) and the velocity (not given) so I still am not getting how the speed of this wave can be found form all of this though.... I am still missing a key variable it seems.
     
  8. Jan 21, 2005 #7

    dextercioby

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    Okay,you're on the right track.2 consecutive maxima determine a wavelength.Question:What is the distance between a maximum and the minimum of same order,distance expressed in terms of the wavelegth...??

    How do you relate that distance to the number (0.029m) given in the problem's text??

    Daniel.
     
  9. Jan 21, 2005 #8
    Since the maximum (crest) and minimum (trough) are equal but opposite, would the distance be inversely proportional?

    1 wavelength / distance

    1/.029

    34.48 m/s
     
  10. Jan 21, 2005 #9

    dextercioby

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    Well,the distance between 2 consecutive maxima is [itex] \lambda [/itex].Since between 2 maxima there is a minimum,and it's located at the middle of the distance,we conlude that the distance between a minimum and the closest maximum is [itex] \frac{\lambda}{2} [/itex]

    Therefore
    [tex]0.029m=\frac{\lambda}{2}[/tex]

    Find lambda,u know the frequency,find the speed...

    Daniel.
     
  11. Jan 21, 2005 #10
    Hi Daniel,

    I see how you figured the consecutive maxima. I was thinking of a maxima as a positive maxima (crest) and a negative maxima (trough). Since it seems that maxima can only be positive then I get how you arrived at wavelength/2.

    To find lambda:

    .029 = wavelength/2 --> wavelength = (.029)(2) = .058

    Then to find speed:

    v = (frequency)(wavelength) --> v = (10700)(.058) = 620.6 m/s

    Would this then be the correct answer?
     
  12. Jan 21, 2005 #11

    dextercioby

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    Looks good.It's big,which means a big pressure and high temperature.Anyway,i think it's the correct answer.

    Daniel.
     
  13. Jan 21, 2005 #12
    Hi Daniel,

    I really appreciate your help! THANK YOU!
     
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